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Standard XII
Mathematics
Question
f
(
x
)
=
|
x
−
1
|
+
|
x
−
3
|
then
f
′
(
2
)
=
0
−
2
1
2
A
2
B
−
2
C
0
D
1
Open in App
Solution
Verified by Toppr
f
(
x
)
=
|
x
−
1
|
+
|
x
−
3
|
=
⎧
⎨
⎩
−
(
x
−
1
)
−
(
x
−
3
)
,
x
≤
1
(
x
−
1
)
−
(
x
−
3
)
,
1
<
x
≤
3
x
−
1
+
(
x
−
3
)
,
x
>
3
⇒
f
(
x
)
=
⎧
⎨
⎩
4
−
2
x
,
x
≤
1
2
,
1
<
x
≤
3
2
x
−
4
,
x
>
3
∴
f
′
(
x
)
=
⎧
⎨
⎩
−
2
,
x
≤
1
0
,
1
<
x
≤
3
2
,
x
>
3
Hence
f
′
(
2
)
=
0
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7
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